Problem: In triangle $ABC$, $AB = 5$, $BC = 4$, and $CA = 3$.

[asy]
defaultpen(1);

pair C=(0,0), A = (0,3), B = (4,0);
draw(A--B--C--cycle);

label("\(A\)",A,N);
label("\(B\)",B,E);
label("\(C\)",C,SW);
[/asy]

Point $P$ is randomly selected inside triangle $ABC$.  What is the probability that $P$ is closer to $C$ than it is to either $A$ or $B$?
Let $\ell$ be the perpendicular bisector of segment $AC$.  We note that the points that are closer to $A$ than they are to $C$ are the points that are on the same side of $\ell$ as $A$. [asy]
defaultpen(1);

pair C=(0,0), A=(0,3), B=(4,0);

pair D = (A+B)/2;
pair E = (C+A)/2;
pair F = (B+C)/2;

pair DH = D + (.5,0);
pair EH = E + (-.5,0);

draw(A--B--C--cycle);
draw(DH--EH,dashed);
fill(E--D--B--C--cycle,gray(.7));

label("\(A\)",A,N);
label("\(B\)",B,(1,0));
label("\(C\)",C,SW);

label("\(\ell\)",DH,(1,0));
label("\(D\)",D,NE);
[/asy]

Since $ABC$ is a 3-4-5 right triangle with a right angle at $C$, $\ell$ is parallel to line $BC$.  Since it passes through the midpoint of $AC$, it also passes through the midpoint of $AB$, which we'll call $D$.

Let $m$ be the perpendicular bisector of segment $BC$.  As before, the points that are closer to $C$ than they are to $B$ are those that lie on the same side of $m$ as $A$, and $m$ also passes through $D$.

[asy]
defaultpen(1);

pair C=(0,0), A=(0,3), B=(4,0);

pair D = (A+B)/2;
pair E = (C+A)/2;
pair F = (B+C)/2;

pair DH = D + (.5,0);
pair EH = E + (-.5,0);
pair DV = D + (0,.5);
pair FV = F + (0,-.5);

draw(A--B--C--cycle);
draw(DV--FV,dashed);
fill(D--F--C--A--cycle,gray(.7));

label("\(A\)",A,N);
label("\(B\)",B,(1,0));
label("\(C\)",C,SW);

label("\(m\)",DV,(0,1));
label("\(D\)",D,NE);
[/asy] Therefore the points that are closer to $C$ than they are to $A$ or $B$ are the points in the shaded rectangle below. [asy]
defaultpen(1);

pair C=(0,0), A=(0,3), B=(4,0);

pair D = (A+B)/2;
pair E = (C+A)/2;
pair F = (B+C)/2;

pair DH = D + (.5,0);
pair EH = E + (-.5,0);
pair DV = D + (0,.5);
pair FV = F + (0,-.5);

draw(A--B--C--cycle);
draw(DV--FV,dashed);
draw(DH--EH,dashed);
fill(D--F--C--E--cycle,gray(.7));

label("\(A\)",A,N);
label("\(B\)",B,(1,0));
label("\(C\)",C,SW);

label("\(m\)",DV,(0,1));
label("\(\ell\)",DH,(1,0));
label("\(D\)",D,NE);
[/asy] The probability we want is then this rectangle's area divided by triangle $ABC$'s area.  There are a few different ways to see that this ratio is $\boxed{\frac{1}{2}}$.  One way is to note that we can divide $ABC$ into 4 congruent triangles, 2 of which are shaded: [asy]
defaultpen(1);

pair C=(0,0), A=(0,3), B=(4,0);

pair D = (A+B)/2;
pair E = (C+A)/2;
pair F = (B+C)/2;

draw(A--B--C--cycle);
fill(D--F--C--E--cycle,gray(.7));

draw(E--D--F);
draw(C--D);

label("\(A\)",A,N);
label("\(B\)",B,(1,0));
label("\(C\)",C,SW);

label("\(D\)",D,NE);
[/asy] Another way is to notice that the rectangle's sides have length $\frac{3}{2}$ and $\frac{4}{2}$, so that the rectangle's area is $\frac{3 \cdot 4}{2 \cdot 2}$.  Since triangle $ABC$'s area is $\frac{3 \cdot 4}{2}$, it follows that the probability we seek is $\boxed{\frac{1}{2}}$, as before.